Codeforces Round #269 (Div. 2) A B C

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先说C

题目链接：http://codeforces.com/problemset/problem/471/C

题目意思：有 n 张卡，问能做成多少种不同楼层（floor）的 house，注意这 n 张卡都要用光。每层 floor 都由一定的 room 构成，每两个相邻 room 规定要有一个公共的ceiling。规定从上到下看，每层 floor 的 room 的数量呈递增的形式排布。

这种东西一般就是看图，先自己从小数开始推算找规律可以发现第i层需要(3*i+2)个，那么前i层总的最少需要就是等差数列求和得(3*i+1)*i/2。因为不能剩余，那么n-(3*i+1)*i/2必须能被3整除，那么从1到(3*i+1)*i/2<=n遍历一下就行 n=10^12 所以O(1e6)时间还是够的

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;int main(){    ll n;    while(~scanf("%I64d",&n))    {        ll ans=0,tmp;        for(ll i=1;(tmp=(3*i+1)*i/2)<=n;i++)        {            if( ( n-tmp )%3 == 0)                ans++;        }        printf("%I64d\n",ans);    }    return 0;}

B, 首先能不能出现三种以上的排列，写代码时用了类似离散化的写法，能的话就随便改两个数的次序就能凑够3种

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;const int MAXN = 2000+20;struct Node{    int id,v;}p[MAXN];int n;bool cmp(Node a, Node b){    return a.v<b.v;}void print(){    printf("%d",p[1].id);    for(int i=2;i<=n;i++)        printf(" %d",p[i].id);    putchar('\n');}void SW(Node &a, Node &b){    Node tmp;    tmp=a;    a=b;    b=tmp;}int main(){    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        {            scanf("%d",&p[i].v);            p[i].id=i;        }        sort(p+1,p+n+1,cmp);        p[n+1].v=-1;        int tmp=1,pre=p[1].v;        ll cnt=1;        int flag=0;        for(int i=2;i<=n+1;i++)            if(pre!=p[i].v)            {                if(tmp>=3){flag=1;break;}                cnt*=tmp;                if(cnt>=3){flag=1;break;}                tmp=1;                pre=p[i].v;            }            else            {                tmp++;            }        if(!flag)puts("NO");        else        {            puts("YES");            print();            int pr=p[1].v,tmp=1;            int cnt=1;            for(int i=2;i<=n+1;i++)///                if(pr!=p[i].v)                {                    if(tmp==2)                    {                        SW(p[i-1],p[i-2]);                        if(cnt<3)print(),cnt++;;                    }                    if(tmp >= 3)                    {                        SW(p[i-1],p[i-2]);                        if(cnt<3) print(),cnt++;                        //cnt++;                        SW(p[i-1],p[i-3]);                        if(cnt<3)print(),cnt++;                        //cnt++;                    }                    if(cnt >=3)break;                    tmp=1;                    pr=p[i].v;                }                else                {                    tmp++;                }        }    }    return 0;}

A 水不过因为flag少写了一个 WA了一次

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;int len[10],vis[12];int main(){    while(~scanf("%d%d%d%d%d%d",len,len+1,len+2,len+3,len+4,len+5))    {        int ans=0;        CL(vis,0);        sort(len,len+6);        int cnt=0,flag=0;        for(int i=0;i<6;i++)        {            vis[len[i]]++;            if(vis[len[i]] == 4)cnt=i,flag=1;        }        if(vis[len[cnt]] == 5)        {            puts("Bear");            continue;        }        if(vis[len[cnt]] == 6)        {            puts("Elephant");            continue;        }        int last=-1,ff=0;        for(int i=0;i<6;i++)            if(i!=cnt)            {                if(vis[len[i]] == 2)ff=1;                if(last==-1){len[0]=len[i];last=1;}                else  len[5]=len[i];            }        if(ff && flag)        {            puts("Elephant");            continue;        }        if(len[0]!=len[5] && flag)        {            puts("Bear");            continue;        }        if(len[0]==len[5] && flag)        {            puts("Elephant");            continue;        }        puts("Alien");    }    return 0;}

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